I was just wondering how you calculate the drop in voltage and indeed if there should be one or should it stay at ~12V but just get weaker.
I'm not sure what you mean by 'getting weaker'. If you mean that it's current-supplying capacity is weaker, then applying an appreciable load would cause the on-load voltage to fall further.
You're asking a complicated question, about the voltage of a rechargeable battery as it discharges. This will vary consdierably according to the type (chemistry) of the battery - whilst the more modern types (NiCad NiMh etc.) tend to maintain their nominal voltage throughout most of their discharge life, and then 'suddenly plummet', the voltage of lead-acid batteries such as you're talking about tends to reduce progressively throughout discharge. It's therefore not something you can 'calculate' directly - you need to look at the relevant 'curves' for the type of battery concerned, and an added complication is that those curves will (as hinted in my recent post) vary a lot depending upon the 'rate of discharge'.
Plus, why, if you jump start it, is it alright for the rest of the day but not the next morning?
I'm not quite sure what you are describing. If you jump start a car which has an essentially flat battery, it will obviously then run (from it's alternator) regardless of the battery. The battery will also get charged to some extent whilst the engine is running, usually enough to start the engine again if it is stopped for a relatively short time. However, if it doesn't get charged up all that much, it might not be man enough to start the engine when it is 'stone cold' the next morning.
Kind Regards, John